abysmaldan abysmaldan - 4 months ago 19
Java Question

Using Recursion to reverse an integer without the use of strings

I have been trying this for some time now but could not get it to work. I am trying to have a method to reverse an integer without the use of strings or arrays. For example, 123 should reverse to 321 in integer form.

My first attempt:

/** reverses digits of integer using recursion */
public int RevDigs(int input)
{
int reverse = 0;
if(input == 0)
{
return reverse;
}
int tempRev = RevDigs(input/10);
if(tempRev >= 10)
reverse = input%10 * (int)Math.pow(tempRev/10, 2) + tempRev;
if(tempRev <10 && tempRev >0)
reverse = input%10*10 + tempRev;
if(tempRev == 0)
reverse = input%10;
return reverse;
}//======================


I also tried to use this, but it seems to mess up middle digits:

/** reverses digits of integer using recursion */
public int RevDigs(int input)
{
int reverse = 0;
if(input == 0)
{
return reverse;
}
if(RevDigs(input/10) == 0)
reverse = input % 10;
else
{
if(RevDigs(input/10) < 10)
reverse = (input % 10) *10 + RevDigs(input/10);
else
reverse = (input % 10)* 10 * (RevDigs(input/10)/10 + 1) + RevDigs(input/10);
}
return reverse;
}


I have tried looking at some examples on the site, however I could not get them to work properly. To further clarify, I cannot use a String, or array for this project, and must use recursion. Could someone please help me to fix the problem. Thank you.

Answer

What about:

public int RevDigs(int input) {
    if(input < 10) {
        return input;
    }
    else {
        return (input % 10) * (int) Math.pow(10, (int) Math.log10(input)) + RevDigs(input/10);
        /* here we:
           - take last digit of input
           - multiply by an adequate power of ten
             (to set this digit in a "right place" of result)
           - add input without last digit, reversed
        */
    }
}

This assumes input >= 0, of course.